Fourier’s Theorem
Fourier’s Theorem
‘any periodic (or regularly repeating) wave,
however complicated, can be described in
terms of an infinite number of sine waves
(of various amplitudes and phases) added
together’
Fourier’s Equation
the sine and cosine
parts deal with phases
of partials
coefficients (give
energies of partials)

v(t )  a0   [an cos( n 1t )  bn sin( n 1t )]
n 1
represents time
add all harmonics
from n=1 to n=
infinity
number of the
harmonic
represents the fundamental
frequency of the waveform
Angular Frequency
1 = angular frequency (in radians)
2
1 
T1
means the
same as:
1
f 
T1
but only works for sine waves
(one cycle of a sine wave in radians = 2π)
Analysis and Synthesis
• Fourier analysis involves taking a wave and
breaking it up into it’s constituent
components
• Fourier synthesis involves constructing a
wave by adding up sine waves
Fourier Analysis/Synthesis
• Fourier analysis and synthesis are used a lot in
D.S.P.
• For example, to apply effects to digital audio
Fourier Synthesis Examples
Creating a Square Wave
1.5
1
v(t)
0.5
0
-0.5
-1
-1.5
0
T/2
time, t
T
3T/2
Coefficient Values for a Square Wave
a0  0
an  0


0,
if
n
is
even


bn   4

, if n is odd 

n

Insert Coefficients
a0  0
an  0
 0, if n is even 


bn   4
, if n is odd 
 n

insert these coefficients:

v(t )  a 0   [a n cos( n 1t )  bn sin( n 1t )]
into Fourier’s
equation:
n 1

to get:
v(t )   bn sin( n 1t ) 
n 1
4
1
1
1

sin

t

sin
3

t

sin
5

t

sin
7

t

.....
1
1
1
1

 
3
5
7
Calculate Terms at Each Time (t)
e.g. at t = T1/4 :
when n=1:
2
1 
T1
sin 1t = sin (2/T1 * T1/4) = sin 2/4
= sin /2 = sin 90 = 1
when n=3:
1/3 sin (31t) = 1/3 sin (3 * 2/T1 * T1/4) = 1/3 sin 6/4
= 1/3 sin 3/2 = 1/3 sin 270 = 1/3 * -1 = -1/3
when n=5:
1/5 sin (51t) = 1/5 sin (5 * 2/T1 * T1/4) = 1/5 sin 10/4
= 1/5 sin 5/2 = 1/5 sin 450 = 1/5 * 1 = 1/5
Calculating Terms Cont…
• according to the formula multiply each
value by 4/
• this makes:
 first term (n= 1) = 1*4/ = 1.273
 second term (n=3) = -1/3 * 4/ = -0.424
 third term (n = 5) = 1/5 * 4/ = 0.255
(at t= T1/4)
(at t = T1/4)
(at t = T1/4)
Plot Values
1.5
1.273
4/ (1/n sin (n1t)
1
0.255
0.5
0
-0.5
-1
-1.5
-0.424
0
T/2
T
time, t
A plot of the first 5 terms (all t values).
3T/2
Add Waves Together
1.5
1
v(t)
0.5
0
-0.5
-1
-1.5
0
T/2
time, t
T
A plot of the first five terms all added together.
3T/2
Add More Terms
1.5
1
v(t)
0.5
0
-0.5
-1
-1.5
0
T/2
time, t
T
A plot of the first twenty terms all added together.
3T/2
Square Wave Freq. Domain Plot
1.4
1.2
amplitude
1
0.8
0.6
0.4
0.2
0
1
31
51
71
frequency
91
111
131
Practicalities
• practical synthesis does not require an
infinite number of sine waves
•
•
DiscoDSPs Vertigo has 256 partials
VirSins Cube has 512
Sawtooth Wave
1.5
1
v(t)
0.5
0
-0.5
-1
-1.5
0
T
time, t
2T
Sawtooth - Freq. Domain Plot
relative value
Coefficients
1.2
a0  0
1
an  0
bn 
0.8
0.6
0.4

0.2
0
1
21
31
41
frequency
51
61
71
2
n
Triangle Wave
1.5
1
v(t)
0.5
0
-0.5
-1
-1.5
0
T
time, t
2T
Triangle – Freq. Domain Plot
Coefficients
0.9
a0  0
0.8
an  0

 0, if n is even 

bn   8

,
if
n
is
odd


n 2 2

0.7
amplitude
0.6
0.5
0.4
0.3
0.2

0.1
0
1
31
51
71
frequency
91
111
131